QJEEM05Ma7014: JEE Main 2005 Mathematics - Class 12 JEEM05Ma7014 Solution

Question

Verified Solution • 2025–26 Session moderate

The value of

\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a &gt; 0,}

is

Accepted Answer

Let

I = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

= \int\limits_{ - \pi }^\pi {{{{{\cos }^2}\left( { - x} \right)} \over {1 + {a^{ - x}}}}} dx

\left[ \, \right.

Using

\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx}

\left. \, \right]

= \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {\alpha ^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

Adding equations

(1)

and

(2)

we get

2I = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\left( {{{1 + {a^x}} \over {1 + {a^x}}}} \right)dx

= \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\,dx

= 2\int\limits_0^\pi {{{\cos }^2}} x\,dx

= 2 \times 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} x\,dx

= 4\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} x\,dx

\Rightarrow I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} \,x\,dx

= 2\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\cos }^2}x\,dx} \right)}

\Rightarrow I\,\, = 2\int\limits_0^{{\pi \over 2}} {dx - 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} } \,x\,dx

\Rightarrow I + I = 2\left( {{\pi \over 2}} \right) = \pi

\Rightarrow I = {\pi \over 2}

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